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CHM 3411 – Problem Set 2 Due date: Wednesday, January 27th Do all of the following problems. Show your work. "I do not like it, and I am sorry I ever had anything to do with it." - Erwin Schrödinger (about quantum mechanics) 1) Find the quantum mechanical operators for the following quantities. a) v (velocity) b) p2 (square of momentum) 2) We discussed the one dimensional particle in a box problem in class. The potential for this system is V(x) = for x 0 or x L V(x) = 0 for 0 < x < L The wavefunctions that are solutions to the TISE for this system and the corresponding energy levels are n(x) = (2/L)1/2 sin(nx/L) inside the box En = n2h2/8mL2 n(x) = 0 outside the box n = 1, 2, 3, … a) Write the wavefunctions corresponding to the n = 2 and n = 3 states of the above system. b) Find P(L/4 < x < L/2), the probability of the particle being found between L/4 and L/2, for the n = 2 and n = 3 states. c) Show that the n = 2 and n = 3 states are orthonormal, that is, show the following 0L 2* 2 dx = 0L 3* 3 dx = 1 ; 0L 2* 3 dx = 0 d) Is f(x) = exp(ikx), where k is a positive constant, an eigenfunction of the Hamiltonian for the particle in a box for the region where V = 0? If so, is it an acceptable solution to the TISE for the system? Justify your answer. e) Find a general formula for the wavelength of light that an electron would need to absorb to move from the n = 1 to the n = 2 state of a particle in a box. Find the value for wavelength (in nm) when the box size is 0.100 nm. 3) The possible values for energy for the particle in a cube are E(nx,ny,nz) = (nx2 + ny2 + nz2) E0 ; where E0 = h2/8mL2 nx = 1, 2, 3, … ny = 1, 2, 3, … (3.1) nz = 1, 2, 3, … Find all of the energy levels for the above system with E 40 E0. List the quantum numbers for each state (nx,ny,nz). Plot your results in an energy level diagram. For each possible value for energy give the degeneracy of the level. Also do the following exercises from Atkins: 9.3 a 9.7 b Solutions 1) a) Since p = mv, v = p/m, and so v = (1/m) (- i d/dx) = - (i/m) d/dx b) p2 = ( - i d/dx)2 = - 2 d2/dx2 2) a) Based on the general formula for the wavefunctions, the n = 2 and n = 3 solutions are (inside the box) n=2 (2/L)1/2 sin(2x/L) n=3 (2/L)1/2 sin(3x/L) b) The general integral we will need is [sin(ax)]2 dx = x/2 - (1/4a) sin(2ax) For n = 2 P = L/4L/2 [(2/L)1/2 sin(2x/L)]2 dx = (2/L) 0L/4 [(sin(2x/L)]2 dx Let a = 2/L, then = (2/L) { x/2 - (L/8) sin(4x/L) }L/4L/2 = (2/L) { (L/8) - (L/8) [sin(2) - sin()] } = 1/4 = 0.2500... For n = 3 P = L/4L/2 [(2/L)1/2 sin(3x/L)]2 dx = (2/L) 0L/4 [(sin(3x/L)]2 dx Let a = 3/L, then = (2/L) { x/2 - (L/12) sin(6x/L) }L/4L/2 = (2/L) { (L/8) - (L/12) [sin (6/2) - sin(3/2)] } = 1/4 - (1/6) 0.1969... c) There are three integrals to consider (since the particle in a box wavefunctions are real, * = ). The general integral needed to do the first two integrals is given in part b of the problem. 0L 2* 2 dx = 0L [(2/L)1/2 sin(2x/L)]2 dx = (2/L) 0L [(sin(2x/L)]2 dx Let a = 2/L, then = (2/L) { x/2 - (L/8) sin(4x/L) }0L = (2/L) { (L/8) - (L/8) sin(4) } = 1 0L 3* 3 dx = 0L [(2/L)1/2 sin(3x/L)]2 dx = (2/L) 0L [(sin(3x/L)]2 dx Let a = 2/L, then = (2/L) { x/2 - (L/8) sin(6x/L) }0L = (2/L) { (L/8) - (L/8) sin(6) } = 1 For the third integral 0L 2* 3 dx = We need the following general integral = sin(mx) sin(nx) dx = sin(m-n)x - sin(m+n)x m2 n2 2(m-n) So 2(m+n) 0L 2* 3 dx = 0L [sin(2x/L) sin(3x/L)] dx Let m = 2/L and n = 3/L, then = { sin(-x/L) - sin(5x/L) }0L (-2/L) (10/L) = 0 , (since sin(-) = sin(5) = sin(0) = 0 ) d) The Hamiltonian, inside the box, is H = - (2/2m) d2/dx2 and so H = - (2/2m) d2/dx2 eikx = - (2/2m) (ik)2 eikx = 2k2/2m eikx so yes, it is an eigenfunction, with eigenvalue = 2k2/2m However it is not an acceptable solution because it cannot satisfy the boundary conditions for the particle in a box, namely, that (0) = (L) = 0. For example, at x = 0 (0) = eik0 = e0 = 1 Since there is no finite normalization constant we can multiply the wavefunction by to make it equal to zero at x = 0, this function is not an acceptable solution to the particle in a box. This is an important point. To be an acceptable solution, a function must be an eigenfunction of H and satisfy all of the boundary conditions for the problem. e) The general formula for the energy is En = n2 E0, where E0 = h2/8mL2. For a transition n = 1 2, the difference in energy is E = E2 - E1 = 4 E0 - E0 = 3 E0. But E = hc/ , and so = hc/E = hc/(3E0) = 8mcL2/3h. For L = 0.100 nm (approximately the size of a hydrogen atom) = 8 (9.11 x 10-31 kg) (2.998 x 108 m/s) (0.100 x 10-9 m)2 = 1.1 x 10-8 m = 11 nm 3 (6.626 x 10-34 J.s) 3) We will first list all of the states systematically, with nx ny nz. nx,ny,nz nx2+ny2+nz2 nx,ny,nz nx2+ny2+nz2 (1,1,1) 3 (4,4,1) 33 (2,1,1) 6 (2,2,2) 12 (3,1,1) 11 (3,2,2) 17 (4,1,1) 18 (4,2,2) 24 (5,1,1) 27 (5,2,2) 33 (6,1,1) 38 (3,3,2) 22 (2,2,1) 9 (4,3,2) 29 (3,2,1) 14 (5,3,2) 38 (4,2,1) 21 (4,4,2) 36 (5,2,1) 30 (3,3,3) 27 (3,3,1) 19 (4,3,3) 34 (4,3,1) 26 (5,3,1) 35 On the next page we give a plot showing the location and numer of all the states with E 40 E0. We then make a few additional comments about this problem. | | | 38 E0. (6,1,1) (1,6,1) (1,1,6) (5,3,2) (5,2,3) (3,5,2) (3,2,5) (2,5,3) (2,3,5) | g=9 | 36 E0. (4,4,2) (4,2,4) (2,4,4) | 35 E0. (5,3,1) (5,1,3) (3,5,1) (3,1,5) (1,5,3) (1,3,5) | 34 E0. (4,3,3) (3,4,3) (3,3,4) | 33 E0. (4,4,1) (4,1,4) (1,4,4) (5,2,2) (2,5,2) (2,2,5) | | | 30 E0. (5,2,1) (5,1,2) (2,5,1) (2,1,5) (1,5,2) (1,2,5) | 29 E0. (4,3,2) (4,2,3) (3,4,2) (3,2,4) (2,4,3) (2,3,4) | | 27 E0. (5,1,1) (1,5,1) (1,1,5) (3,3,3) | 26 E0. (4,3,1) (4,1,3) (3,4,1) (3,1,4) (1,4,3) (1,3,4) | | 24 E0. (4,2,2) (2,4,2) (2,2,4) | | 22 E0. (3,3,2) (3,2,3) (2,3,3) | 21 E0. (4,2,1) (4,1,2) (2,4,1) (2,1,4) (1,4,2) (1,2,4) | | 19 E0. (3,3,1) (3,1,3) (1,3,3) | 18 E0. (4,1,1) (1,4,1) (1,1,4) | 17 E0. (3,2,2) (2,3,2) (2,2,3) | | | 14 E0. (3,2,1) (3,1,2) (2,3,1) (2,1,3) (1,3,2) (1,2,3) | | 12 E0. (2,2,2) | 11 E0. (3,1,1) (1,3,1) (1,1,3) | | 9 E0. (2,2,1) (2,1,2) (1,2,2) | | | 6 E0. (2,1,1) (1,2,1) (1,1,2) | | | 3 E0. (1,1,1) | | . We can count the number of states in a particular energy range 36-40 E0. 31-35 E0. 26-30 E0. 21-25 E0. 16-20 E0. 11-15 E0. 6-10 E0. 1-5 E0. 12 states 15 states 22 states 12 states 9 states 10 states 6 states 1 state These data are plotted below as # of states vs E1/2. g=3 g=6 g=3 g=6 g=6 g=6 g=4 g=6 g=3 g=3 g=6 g=3 g=3 g=3 g=6 g=1 g=3 g=3 g=3 g=1 It can be shown that for sufficiently large values of E that # of states ~ E 1/2. Exercise 9.3 a For the n = 1 state of the particle in a box (x) = (2/L)1/2 sin(x/L) <p> = 0L [(2/L)1/2 sin(x/L)] (-i d/dx) [(2/L)1/2 sin(x/L)] dx = ( -i) (2/L) (/L) 0L sin(x/L) cos(x/L) dx The general form for this integral is sin(ax) cos(ax) dx = (1/2a) [sin(ax)]2 Let a = /L, then = - (2i/L2) (L2/2) [sin(x/L]0L = 0 <p2> = 0L [(2/L)1/2 sin(x/L)] (-2 d2/dx2) [(2/L)1/2 sin(x/L)] dx = ( -2) (2/L) (/L)2 (-1) 0L [sin(x/L)]2 dx The general form for this integral has been given in a previous problem. Let a = /L, then = (222/L3) { x/2 - (L/4) sin(2x/L }0L = (222/L3) (L/2) = 22/L2 = h2/4L2 Exercise 9.7b The problem is worded somewhat ambiguously, but it is saying that E = n2 E0, where E0 = h2/8mL2 m(N2) = (28.0 x 10-3 kg/mol) (1 mol/6.022 x 1023) = 4.65 x 10-26 kg So E0 = (6.626 x 10-34 J.s)2 = 1.18 x 10-42 J 8 (4.65 x 10-26 kg) (1.00 m)2 Since E = (3/2) kT = (3/2) (1.381 x 10 -23 J/K) (300. K) = 6.21 x 10-21 J Then n2 = E/E0 = (6.21 x 10-21)/(1.18 x 10-42) = 5.26 x 1021 n = (n2)1/2 = 7.25 x 1010 The separation between adjacent energy levels is E = E(n+1) - E(n) = (n+1)2E0 - n2E0 = [(n2 + 2n + 1) - n2] E0 = (2n+1) E0 For the above value of n, E = [2 (7.25 x 1010) + 1] (1.18 x 10-42 J) = 1.71 x 10-31 J Note that since E is small compared to E the energy levels are essenially continuous. Finally, the de Broglie wavelength is deBroglie = h/mv But E = p2/2m, and so p = mv = (2mE)1/2 = [2 (4.65 x 10-26 kg) (6.21 x 10-21 J)]1/2 = 2.40 x 10-23 kg.m/s deBroglie = h/mv = (6.626 x 10-34 J.s)/(2.40 x 10-23 kg.m/s) = 2.76 x 10-11 m Particles behave classically (that is, as particles) on size scales large compared to the de Broglie wavelength. Since the de Broglie wavelength is so small the particle will behave classically.